3.21.7 \(\int \frac {1}{(d+e x)^{5/2} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\) [2007]

Optimal. Leaf size=153 \[ \frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}-\frac {2 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}} \]

[Out]

2/5/(-a*e^2+c*d^2)/(e*x+d)^(5/2)+2/3*c*d/(-a*e^2+c*d^2)^2/(e*x+d)^(3/2)-2*c^(5/2)*d^(5/2)*arctanh(c^(1/2)*d^(1
/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/(-a*e^2+c*d^2)^(7/2)+2*c^2*d^2/(-a*e^2+c*d^2)^3/(e*x+d)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 53, 65, 214} \begin {gather*} -\frac {2 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}+\frac {2 c^2 d^2}{\sqrt {d+e x} \left (c d^2-a e^2\right )^3}+\frac {2 c d}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

2/(5*(c*d^2 - a*e^2)*(d + e*x)^(5/2)) + (2*c*d)/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) + (2*c^2*d^2)/((c*d^2 -
a*e^2)^3*Sqrt[d + e*x]) - (2*c^(5/2)*d^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*
d^2 - a*e^2)^(7/2)

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx &=\int \frac {1}{(a e+c d x) (d+e x)^{7/2}} \, dx\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {(c d) \int \frac {1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{c d^2-a e^2}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {\left (c^2 d^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{\left (c d^2-a e^2\right )^2}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}+\frac {\left (c^3 d^3\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{\left (c d^2-a e^2\right )^3}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}+\frac {\left (2 c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {c d^2}{e}+a e+\frac {c d x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e \left (c d^2-a e^2\right )^3}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}-\frac {2 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 145, normalized size = 0.95 \begin {gather*} \frac {6 a^2 e^4-2 a c d e^2 (11 d+5 e x)+2 c^2 d^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )}{15 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {2 c^{5/2} d^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

(6*a^2*e^4 - 2*a*c*d*e^2*(11*d + 5*e*x) + 2*c^2*d^2*(23*d^2 + 35*d*e*x + 15*e^2*x^2))/(15*(c*d^2 - a*e^2)^3*(d
+ e*x)^(5/2)) - (2*c^(5/2)*d^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(-(c*d^2)
+ a*e^2)^(7/2)

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Maple [A]
time = 0.74, size = 146, normalized size = 0.95

method	result	size
derivativedivides	\(-\frac {2 c^{3} d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}-\frac {2}{5 \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {2 c d}{3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}\)	\(146\)
default	\(-\frac {2 c^{3} d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}-\frac {2}{5 \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {2 c d}{3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x,method=_RETURNVERBOSE)

[Out]

-2*c^3*d^3/(a*e^2-c*d^2)^3/((a*e^2-c*d^2)*c*d)^(1/2)*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))-2/5/(
a*e^2-c*d^2)/(e*x+d)^(5/2)-2/(a*e^2-c*d^2)^3*c^2*d^2/(e*x+d)^(1/2)+2/3/(a*e^2-c*d^2)^2*c*d/(e*x+d)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (128) = 256\).
time = 4.29, size = 777, normalized size = 5.08 \begin {gather*} \left [-\frac {15 \, {\left (c^{2} d^{2} x^{3} e^{3} + 3 \, c^{2} d^{3} x^{2} e^{2} + 3 \, c^{2} d^{4} x e + c^{2} d^{5}\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d x e + 2 \, c d^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {x e + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} - a e^{2}}{c d x + a e}\right ) - 2 \, {\left (35 \, c^{2} d^{3} x e + 23 \, c^{2} d^{4} - 5 \, a c d x e^{3} + 3 \, a^{2} e^{4} + {\left (15 \, c^{2} d^{2} x^{2} - 11 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}}{15 \, {\left (3 \, c^{3} d^{8} x e + c^{3} d^{9} - a^{3} x^{3} e^{9} - 3 \, a^{3} d x^{2} e^{8} + 3 \, {\left (a^{2} c d^{2} x^{3} - a^{3} d^{2} x\right )} e^{7} + {\left (9 \, a^{2} c d^{3} x^{2} - a^{3} d^{3}\right )} e^{6} - 3 \, {\left (a c^{2} d^{4} x^{3} - 3 \, a^{2} c d^{4} x\right )} e^{5} - 3 \, {\left (3 \, a c^{2} d^{5} x^{2} - a^{2} c d^{5}\right )} e^{4} + {\left (c^{3} d^{6} x^{3} - 9 \, a c^{2} d^{6} x\right )} e^{3} + 3 \, {\left (c^{3} d^{7} x^{2} - a c^{2} d^{7}\right )} e^{2}\right )}}, -\frac {2 \, {\left (15 \, {\left (c^{2} d^{2} x^{3} e^{3} + 3 \, c^{2} d^{3} x^{2} e^{2} + 3 \, c^{2} d^{4} x e + c^{2} d^{5}\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {x e + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d x e + c d^{2}}\right ) - {\left (35 \, c^{2} d^{3} x e + 23 \, c^{2} d^{4} - 5 \, a c d x e^{3} + 3 \, a^{2} e^{4} + {\left (15 \, c^{2} d^{2} x^{2} - 11 \, a c d^{2}\right )} e^{2}\right )} \sqrt {x e + d}\right )}}{15 \, {\left (3 \, c^{3} d^{8} x e + c^{3} d^{9} - a^{3} x^{3} e^{9} - 3 \, a^{3} d x^{2} e^{8} + 3 \, {\left (a^{2} c d^{2} x^{3} - a^{3} d^{2} x\right )} e^{7} + {\left (9 \, a^{2} c d^{3} x^{2} - a^{3} d^{3}\right )} e^{6} - 3 \, {\left (a c^{2} d^{4} x^{3} - 3 \, a^{2} c d^{4} x\right )} e^{5} - 3 \, {\left (3 \, a c^{2} d^{5} x^{2} - a^{2} c d^{5}\right )} e^{4} + {\left (c^{3} d^{6} x^{3} - 9 \, a c^{2} d^{6} x\right )} e^{3} + 3 \, {\left (c^{3} d^{7} x^{2} - a c^{2} d^{7}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

[-1/15*(15*(c^2*d^2*x^3*e^3 + 3*c^2*d^3*x^2*e^2 + 3*c^2*d^4*x*e + c^2*d^5)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*
x*e + 2*c*d^2 + 2*(c*d^2 - a*e^2)*sqrt(x*e + d)*sqrt(c*d/(c*d^2 - a*e^2)) - a*e^2)/(c*d*x + a*e)) - 2*(35*c^2*
d^3*x*e + 23*c^2*d^4 - 5*a*c*d*x*e^3 + 3*a^2*e^4 + (15*c^2*d^2*x^2 - 11*a*c*d^2)*e^2)*sqrt(x*e + d))/(3*c^3*d^
8*x*e + c^3*d^9 - a^3*x^3*e^9 - 3*a^3*d*x^2*e^8 + 3*(a^2*c*d^2*x^3 - a^3*d^2*x)*e^7 + (9*a^2*c*d^3*x^2 - a^3*d
^3)*e^6 - 3*(a*c^2*d^4*x^3 - 3*a^2*c*d^4*x)*e^5 - 3*(3*a*c^2*d^5*x^2 - a^2*c*d^5)*e^4 + (c^3*d^6*x^3 - 9*a*c^2
*d^6*x)*e^3 + 3*(c^3*d^7*x^2 - a*c^2*d^7)*e^2), -2/15*(15*(c^2*d^2*x^3*e^3 + 3*c^2*d^3*x^2*e^2 + 3*c^2*d^4*x*e
+ c^2*d^5)*sqrt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(x*e + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*x
*e + c*d^2)) - (35*c^2*d^3*x*e + 23*c^2*d^4 - 5*a*c*d*x*e^3 + 3*a^2*e^4 + (15*c^2*d^2*x^2 - 11*a*c*d^2)*e^2)*s
qrt(x*e + d))/(3*c^3*d^8*x*e + c^3*d^9 - a^3*x^3*e^9 - 3*a^3*d*x^2*e^8 + 3*(a^2*c*d^2*x^3 - a^3*d^2*x)*e^7 + (
9*a^2*c*d^3*x^2 - a^3*d^3)*e^6 - 3*(a*c^2*d^4*x^3 - 3*a^2*c*d^4*x)*e^5 - 3*(3*a*c^2*d^5*x^2 - a^2*c*d^5)*e^4 +
(c^3*d^6*x^3 - 9*a*c^2*d^6*x)*e^3 + 3*(c^3*d^7*x^2 - a*c^2*d^7)*e^2)]

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Sympy [A]
time = 10.23, size = 141, normalized size = 0.92 \begin {gather*} - \frac {2 c^{2} d^{2}}{\sqrt {d + e x} \left (a e^{2} - c d^{2}\right )^{3}} - \frac {2 c^{2} d^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}} \left (a e^{2} - c d^{2}\right )^{3}} + \frac {2 c d}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e^{2} - c d^{2}\right )^{2}} - \frac {2}{5 \left (d + e x\right )^{\frac {5}{2}} \left (a e^{2} - c d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)

[Out]

-2*c**2*d**2/(sqrt(d + e*x)*(a*e**2 - c*d**2)**3) - 2*c**2*d**2*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d
)))/(sqrt((a*e**2 - c*d**2)/(c*d))*(a*e**2 - c*d**2)**3) + 2*c*d/(3*(d + e*x)**(3/2)*(a*e**2 - c*d**2)**2) - 2
/(5*(d + e*x)**(5/2)*(a*e**2 - c*d**2))

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Giac [A]
time = 1.36, size = 211, normalized size = 1.38 \begin {gather*} \frac {2 \, c^{3} d^{3} \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} c^{2} d^{2} + 5 \, {\left (x e + d\right )} c^{2} d^{3} + 3 \, c^{2} d^{4} - 5 \, {\left (x e + d\right )} a c d e^{2} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )}}{15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

2*c^3*d^3*arctan(sqrt(x*e + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 -
a^3*e^6)*sqrt(-c^2*d^3 + a*c*d*e^2)) + 2/15*(15*(x*e + d)^2*c^2*d^2 + 5*(x*e + d)*c^2*d^3 + 3*c^2*d^4 - 5*(x*
e + d)*a*c*d*e^2 - 6*a*c*d^2*e^2 + 3*a^2*e^4)/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*(x*e +
d)^(5/2))

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Mupad [B]
time = 0.14, size = 171, normalized size = 1.12 \begin {gather*} -\frac {\frac {2}{5\,\left (a\,e^2-c\,d^2\right )}+\frac {2\,c^2\,d^2\,{\left (d+e\,x\right )}^2}{{\left (a\,e^2-c\,d^2\right )}^3}-\frac {2\,c\,d\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}}{{\left (d+e\,x\right )}^{5/2}}-\frac {2\,c^{5/2}\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(5/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)),x)

[Out]

- (2/(5*(a*e^2 - c*d^2)) + (2*c^2*d^2*(d + e*x)^2)/(a*e^2 - c*d^2)^3 - (2*c*d*(d + e*x))/(3*(a*e^2 - c*d^2)^2)
)/(d + e*x)^(5/2) - (2*c^(5/2)*d^(5/2)*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^3*e^6 - c^3*d^6 + 3*a*c^2*d^4*
e^2 - 3*a^2*c*d^2*e^4))/(a*e^2 - c*d^2)^(7/2)))/(a*e^2 - c*d^2)^(7/2)

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